TS EAMCET 10 May 2024 Shift 2 Solved Paper

Given data indicates: $\;r_1+r_2=3R$ $\;r_2+r_3=2R$ Let's analyze these conditions: From $r_1+r_2=3R$ : We start by substituting in the formula: $\;\frac{\Delta}{s-a}+\;\frac{\Delta}{s-b}=\;\frac{3abc}{4\Delta}$ Simplifying, we have: $\Delta^2\left(\;\frac{1}{s-a}+\;\frac{1}{s-b}\right)=\;\frac{3abc}{4}$ Using the identity: $s(s-b)(s-c)+s(s-a)(s-c)=\;\frac{3abc}{4}$ Combining terms: $s(s-c)(s-b+s-a)=\;\frac{3abc}{4}$ Further simplification: $s(s-c)c=\;\frac{3abc}{4}$ Resulting relationship: $s(s-c)=\;\frac{3ab}{4}$ Therefore: $\sqrt{\;\frac{s(s-c)}{ab}}=\;\frac{\sqrt{3}}{2}$ Which implies: $\cos\left(\;\frac{C}{2}\right)=\;\frac{\sqrt{3}}{2}$ Hence, $\;\frac{C}{2}=30^{\circ}\;\text{ or }\;C=60^{\circ}$ From $r_2+r_3=2R$ : Use: $s(s-a)=\;\frac{2bc}{4}$ Gives: $\sqrt{\;\frac{s(s-a)}{bc}}=\;\frac{1}{\sqrt{2}}$ Therefore: $\cos\left(\;\frac{A}{2}\right)=\;\frac{1}{\sqrt{2}}$ Thus, $\;\frac{A}{2}=45^{\circ}\;\text{ or }\;A=90^{\circ}$Final Angles of the Triangle: $\;\angle A=90^{\circ}$ $\;\angle B=60^{\circ}$ $\;\angle C=30^{\circ}$ This implies the triangle does not fit typical ratios like $a:b:c=2:1:\sqrt{3}$ ; instead, the lengths of sides must be inconsistent, i.e., $a \neq b \neq c$ , which corresponds to a non-equilateral, non-isosceles right triangle.